Only 17% of all 64-bit Integers are products of two 32-bit integers | Hacker News Reader

Only 17% of all 64-bit Integers are products of two 32-bit integers

lemire.me

226 pointssebg103 comments

> I find it interesting to consider that if you pick a value at random, it will usually fail! That is, most 64-bit integers cannot be written as the product of two 32-bit integers.

While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already, and considering even a tiny amount of overlapping results gets you there.

What gets interesting is actually trying to quantify the overlapping results.

A lot of the remaining is multiples of 4, which you can either get from having a 2 in both factors or a 4 in one (multiples of 9 are similar).

... or just considering the even numbers almost all of them are 2 x N where N>2^32 and that gets you to within a hair of "most" and if you add in the odd thirds for which the same is true you get a bound of 2/3 - epsilon.

It's a bit more subtle than that -- most n>2^32 are not prime in which case 2 x n has more factorizations you would have to check.

(Just by way of example, for n=2^33, 2n=2^34 but also =2^17*2^17)

I'm not sure I understand your comment. Just because a number of of the form 2N with N > 2^32 doesn't mean it can't also be written as the product of two numbers below 2^32.
E.g. 2^40 = 2^20 * 2^20

All the primes above 2^32 are out, but that accounts for only two point something percent.

But also all of their multiples. I suspect that those account for the vast majority.

Each x is prime with probability 1/ln(x), each x has M/x multiples less than M, as a fraction of M that is just 1/x. Together that makes 1/(x ln(x)) with the indefinite integral ln(ln(x)). If we plug in 2^32 and 2^64 [1], we get ln(2). So about 69.3 % of all 64 bit integers should have a prime factor larger than 2^32 and therefore not be the product of two 32 bit integers. That leaves about 13 % unaccounted. Three prime factors all larger than 2^32/2, five prime factors all larger than 2^32/3, and so on cannot be packed into two 32 bit integers. Not sure to how much this will add up.

[1] The bounds are important because they guarantee that there is at most one prime factor from that range and this ensures that we are not double counting anything. If the upper bound was larger than the square of the lower bound, then we would have to worry about double counting numbers with more than one large prime factor.

Nice work. I guess "vast majority" is overstating the case, but majority anyway.

The simple explanation is just that lot of ways that numbers can multiply to produce the same product

Like an odd number x times an even number y, x* y produces the same product as x* 2 and y/2

Same for a multiple of 3 c and and a non multiple of 3 d, c * d = c/3 * d*3

This is just looking at it from a different perspective. Both, one 64 bit integer and the product of two 32 bit integers represent a number up to 2^64 with 64 bits. But while all 64 bit integers are unique, there are, as you say, several representation for some numbers as the product of two 32 bit integers and therefore it is impossible to represent all 64 bit integers. Commutativity alone costs you about 50 percent of all numbers in the range as x * y and y * x represent the same 64 bit number but with two different representation as a product of two 32 bit numbers, at least if x and y are different. But this tells you nothing about the numbers that you can not represent, only that they must exist. I was looking at it from this other perspective, which numbers are not representable as a product and why.

> While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already

It's much worse than that. It's difficult for a 64-bit product to have the high bit set if the multiplicands are both no larger than 32 bits.

I wouldn't say difficult. To set the high bit, the geometric mean of the two multiplicands has to be at least 3 billion. 32 bit numbers go up to 4.3 billion so that's quite common.

If I did the correct integral (the area inside the unit square above y=.5/x), then 15.34% of products should have the high bit set. That's much less than half but it's still happening constantly.

> Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63.

Not sure I understand.

Adding two 32 bit integers takes you to 33 bit integers. (1111 + 1111 = 11110).

Addition doesn't care about order, so you're instantly cutting 2^33 possibilities down to 2^32. Or so is your argument. But in reality you can reach nearly all of those 2^33 numbers.

Addition in this case is cutting from 2^64 to 2^33-1.

The 2^64 number is the number of inputs. For an operation which is commutative, you expect the outputs to be 2^63+2^32 or smaller, since you’ve introduced symmetry.

Concatenating arbitrary 32 bit ints covers all possible 64 bit ints. So the space of all pairs of 32 bit ints is in bijection with 64 bit ints.

Commutativity introduces a relation on pairs of 32 bit ints (a,b) ~ (b,a), which accounts for one bit of information. Thus, at most 50% of 64bit ints show up as products of 32 bit ints.

Ah, fair enough, thanks everyone. So basically the argument is if that we have a deterministic function taking a pair (x_1, x_2) with x_i in X with |X| = M, then the function can produce at most M^2 outputs. And knowing that the function is symmetric cuts it down to M(M+1)/2. (Which is still far bigger than the 2M in my addition analogy.) Cheers.

Except the perfect squares don't reduce by half, so it's not quite 50% but it's very close.

Ha, fair!

Some of them, with notable exception of perfect squares of primes, can be expressed by products of different combinations of factors.

E.g., 6^2 = (223)3 = 2(233).

One of ~22 (ln(2^32)) perfect squares will be a square of perfect prime. Most won't.

Decoding this so it's readable: 6^2 = (2*2*3)*3 = 2*(2*3*3).

(You can escape * with \: \*)

The 2^64 in gps argument comes from the number of pairs of 32 bit numbers, not from the upper bound of multiplying two 32 bit numbers. So for the addition case the symmetry argument is still only good enough to get you down to about 2^63, which doesn't help you at all because you have much stronger information from the upper bound.

Yeah the number sounds a lot less impressive if you say that you only get 2^61.44 integers out of 2^64. In other words, a 4% entropy loss.

Information quantities are more meaningfully expressed in number of bits.

Exactly. Being precise about logarithmic vs linear utilizations is key here. I tried making a similar point about the inefficiency of IEEE-754 redundant NaN encodings here: https://arxiv.org/pdf/2508.05621

Having ~a quadrillion redundant bitstrings all mapping to NaN sounds pretty bad, but logarithmic/information utilization-wise, this is actually not too bad.

I've been looking at the 8087 NaN circuitry lately. Having 2^53 (or whatever) values for NaN was supposedly a feature: "the large number of NAN values that are available, provide the sophisticated programmer with a tool that can be applied to a variety of special situations." For example, the different NaN values could hold debugging information to track down errors.

See "8087 Numeric Data Processor" page S-74: https://ethw.org/w/images/2/2f/Intel_8086_family_users_numer...

You'll see some scripting languages (ab)use this. Where the native "number" type is a 64 bit float and only one NaN bit pattern is a real NaN. The others smuggle a pointer to an object in the lower bits. This way you don't spend any memory overhead indicating if a given variable contains a primitive or an object.

> For example, the different NaN values could hold debugging information to track down errors.

Unfortunately IEEE didn't bother specifying NaN propagation semantics so it ended up pretty useless.

Why does order matter?

Whether a 64-bit number can be written as the product of two 32-bit ones depends only on the prime factors of the 64-bit number - it's a property of the number itself, and apparently 17% of 64-bit numbers have this property.

The input space is 32 + 32 = 64 bits. The output space is 64 bits. So the best you can do is an 1-to-1 mapping.

However, since a * b = b * a, our input space has a lot of duplicate outputs. So from this alone you can conclude roughly half of the output space must be uncovered by any input pair, simply because there aren't enough input pairs.

OK - thanks. I must have misunderstood what the other poster was saying, since I thought they were objecting to the "most" characterization.

I wasn't saying it's wrong, I was saying that "most" is so easy to reach that it's a trivial and rather boring threshold.

Ok so I think I understand your insight: the number of 64 bit numbers you can get from multiplying two 32 bit numbers is the number of distinct results. I guess it follows that, of those 64 bit integers that can be written as the product of 2 32 bit ints, on average they can be factored into 32 bit ints 6 different ways. The only ones that could possibly be written as such a product exactly one way are the prefect squares.

The word “most” is indeed not very surprising if taken to mean >50% (as you do), but the surprising fact (admittedly not relevant for 2^64 in the quoted sentence) is that “most” can be replaced with “almost all” in the technical sense meaning “tending to 1” i.e. “1 - o(1)”, i.e. “eventually greater than 1-epsilon for any epsilon”.

Here's the multiplication table up to 9 (so for n=10 in place of n=2^64), and it already contains only 37 distinct products among its 100 entries:

     × |  0  1  2  3  4  5  6  7  8  9
    ---+------------------------------
     0 |  0  0  0  0  0  0  0  0  0  0
     1 |  0  1  2  3  4  5  6  7  8  9
     2 |  0  2  4  6  8 10 12 14 16 18
     3 |  0  3  6  9 12 15 18 21 24 27
     4 |  0  4  8 12 16 20 24 28 32 36
     5 |  0  5 10 15 20 25 30 35 40 45
     6 |  0  6 12 18 24 30 36 42 48 54
     7 |  0  7 14 21 28 35 42 49 56 63
     8 |  0  8 16 24 32 40 48 56 64 72
     9 |  0  9 18 27 36 45 54 63 72 81

That is, in decimal, only 37% of (up to) two-digit numbers can be written as products of two one-digit numbers. This fraction, which drops to 28% at n=100, only drops to 17% at n=2^64 (per the article). So it decreases VERY slowly, and it's nontrivial that it actually goes to 0.

It's an interesting fact, but it's weakened a lot by happening so slowly and not matching the everyday definition of "almost all".

And it's weakened even more by realizing that while you can get the raw fraction as low as you want, shrinking your list of products by n digits requires numbers with an exponential number of digits.

I think your second paragraph is the same as “so slowly”, just quantified. And I'd argue that the technical meaning of “almost all” does match the everyday definition: if you told someone that “for sufficiently large N, almost all 2N-bit numbers are not the product of two N-bit numbers”, they'd probably think of “almost all” as a fraction like 99.99% or 99.999999% or whatever, and whatever fraction they picked, the statement would be true (with the threshold for “sufficiently large” depending on the fraction they picked).

So whether “only 17%” is interesting or not depends on whether you see it as a stand-in for “less than half”, or “a number close to 0”.

(Posting this comment mainly to correct an error in my previous comment: in both places that I wrote “n=2^64” I should have instead written “n=2^32”.)

> if you told someone that “for sufficiently large N, almost all 2N-bit numbers are not the product of two N-bit numbers”, they'd probably think of “almost all” as a fraction like 99.99% or 99.999999% or whatever, and whatever fraction they picked, the statement would be true (with the threshold for “sufficiently large” depending on the fraction they picked).

I think people would agree with that, yes. But that's a significantly weaker claim than your original one. The original version was "tends to 1 at large n" = "almost all", but this version is that once you reach large n it's "almost all". These different tests give completely different answers for the numbers you'd ever actually use.

And entirely separate from that, if you laid it out as "for 8 million* digit numbers, only one in a billion are products of 4 million digit numbers, so only enough to fill out 7,999,991 digits", I don't know if that really qualifies for "almost all" anymore. The fraction of hits is important, but so is the fraction of digits and entropy, and as you make the numbers bigger you approach 0.0% loss of digits and entropy.

* Placeholder number, I did not do the actual calculation here.

There are about 4 billion 64 bit integers for each 32 bit integer.

The chance of a random 64 bit integer being a 32 bit integer is 0.0000000233 %

The chance of a random 64 bit integer being a product of two 32 bit integers is 17%

Nice

There are about 18.446 quintillion more 64-bit integers than 32-bit integers.

I think they meant to write "There are about 4 billion TIMES more 64 bit integers than 32 bit integers".

Indeed, edited the mistake

True, but there are as many 64-bit integers as pairs of 32-bit integers.

Therefore the fact that relatively few 64-bit numbers are products of 32-bit integers means that a lot of pairs of 32-bit integers give by multiplication the same product.

That seems intuitively true given that most 32-bit numbers are composite, so if you have

X = ab and aY < 2^32 and bY < 2^32:

X × Y = X/a × aY = X/b × bY = Y × X = aY × X/a = bY × X/b

Which is 6 pairs resulting in the same product. This will be reduced if e.g. aY = X, but still...

There are about 2^64 more 64-bit integers than 32-bit integers.

The chance of a random 64-bit integer matching some pair of 32-bit integers is a 100%, though.

Or, the odds of a random 64-bit integer being a 32-bit integer are the same as you or me guessing a random 32 bit integer.

Wonder what the limit is as you add more 32 bit integers to the product. Just the primes over 32 bit?

If you're allowed to multiply as many 32-bit numbers as you want, the only numbers you won't be able to achieve by so doing are those with any prime factor larger than 2^32.

This is more than just the prime numbers. For example, a 41-bit prime can be multiplied by 16 and it will still fit into 64 bits.

What are you assuming about overflow? Three 32-bit numbers multiply out to 96 bits.

If you bring overflow into the mix things become a lot more complicated. You likely don't even need 32 bits, the numbers 2 and 3 might be enough (I don't know for sure or if there's a quick way to check).

Well, if you "bring overflow into the mix", what you get depends on your behavior when overflowing.

If you say that you want to be doing modular arithmetic instead of arithmetic, it doesn't look like 2 and 3 are enough. You're looking for a solution to

    2ᵃ * 3ᵇ ≡ n (mod 2⁶⁴)

If n is even, we can supply any number of 2 factors by fiddling with a. We can assume without loss of generality that n is odd and a = 0. Now we want

    3ᵇ ≡ n (mod 2⁶⁴)

for odd n.

If I'm reading wikipedia correctly, we know that this will fail for some n:

https://en.wikipedia.org/wiki/Primitive_root_modulo_n

> In symbols, g is a primitive root modulo n if for every integer a coprime to n, there is some integer k for which gᵏ ≡ a (mod n).

This is what we want, with g = 3, k = b, a = n₁, and n₂ = 2⁶⁴. Our restriction that n₁ (our n) is odd satisfies the requirement that a be coprime to n₂ (wikipedia's n, the modulus).

The article continues:

> a primitive root exists modulo n if and only if n is 4, pᵏ or 2pᵏ for some odd prime number p and some k ≥ 0.

2⁶⁴ does not satisfy this requirement and therefore there is no primitive root modulo 2⁶⁴. As such, 3 is not a primitive root modulo 2⁶⁴.

Interesting, one of those situations where the odd primes behave slightly differently. Which means it's probably above my paygrade.

I did say things would get complicated.

> the proportion of all 2n-bit values that can be generated by the product of two n-bit values goes to zero as n becomes large. This means that if you have, say, 10000000-bit integers multiplying 10000000-bit integers, you’d expect relatively few 100000000000000-bit integers to be produced.

That should be "relatively few 20000000-bit integers", right?

Perhaps it's binary.

It looks like that has been fixed in the article. Good catch!

I dream of a future where all 64-bit integers are products of 32-bit integers. Together, we can change math for the better.

I upvoted you, not because I think your joke is particularly great, but I hate that HN has this tendency to downvote comments that are clearly meant as a humorous contribution. And I get it, no-one wants HN to turn into Reddit. I also understand that not every joke lands. But I just think it's unnecessary to downvote, you could simply ignore.

"Ignore" is one of those things that sounds like it's a neutral choice but really isn't in practice - it's still just saying "can only ever be positively pressured". IMO people shouldn't go as far as flag though, at the very least, and if it's already at the bottom of the sort there is no sense dumping on it further.

My current comment itself, for instance, also doesn't really add anything to the discussion about the article and I'd have no expectation people leave it from going negative. Maybe the will, maybe they won't, but there is no reason to expect they should in principle of me loving tangents :D.

Why stop there? We can dream of a future where math is bent to our will [0] for the betterment of all mankind!

0: https://en.wikipedia.org/wiki/Indiana_pi_bill

1 + 1 = 3 (for sufficiently large values of 1)

It helps if you take the limit of 1 going towards 1.5.

Most 1s won't go towards 1.5, but sometimes you're lucky.

I thought you were making a joke but if we're assuming that the 1's are being rounded or truncated before the final value cake is produced I guess you are right.

There should be a law!

Indeed, but justice requires that we recursively continue all the way to the base case, until all 32-bit integers are products of 16-bit integers, all 16-bit integers are products of 8-bit integers, all 8-bit integers are products of 4-bit integers, all 4-bit integers are products of 2-bit integers, and all 2-bit integers are products of 1-bit integers. Only when we have reach all the way down that list to the very, very smallest of the numbers around us and brought justice to them will the future be able to arrive. I literally can not wait for that day.

Enough of this divided binary world, we are all one

Did you just misdigit me

That would require multiplication to be non-commutative, right?

Maybe we can reach there by using integers as fixed point decimals?

Cryptographers hate this trick

This feels like a underlying property that contributes to of Benford's Law[0]. That is, most numbers we measure and record are the results of various independent (addition) and dependent (multiplication) factors stacking together, and we observe this property in the distribution of them.

[0]: https://en.wikipedia.org/wiki/Benford%27s_law

> You might be able to come up with a more efficient algorithm.

Challenge accepted. Suppose we want to know the answer to 3 decimal places (so we'd match the headline). And suppose I allow my algorithm to be wrong one in a thousand times ("probably approximately correct").

Then sample some constant number C of random 64 bit integers. Run the following algorithm which separates each random sample into one of three classes: Y (has 32 but factors), N (does not have 32 bit factors), U (unknown).

Check if prime using probabilistic miller rabin. (Error prob goes to zero exponentially fast). If prime, return N. If it's not a prime, then run T steps of pollard rho to determine whether the number has 32 but factors; return Y,N, or U depending of the factors found up to step T.

The key observation is that T can be chosen to make the UNKNOWN class very small (with high probability), and so our estimate should rapidly converge to 17%Y, 83%N, ~0.001%U

For fixed error tolerance, this would run in roughly a constant number of iterations, independent of N.

Even if you have 32-bit factors the number may not be the product of two 32-bit numbers. For example 2^62*3 cannot be split as either (2^32, 2^30*3) or (2^31, 2^31*3). In both cases one factor does not fit in 32 bits.

Ah, yes, this is true, but it's not really a counterexample, since this would show up in the N or U bucket. But I think the issue is that my sketch algorithm is, well, pretty sketchy.

Working on coding it up... it converges to 17±0.5%for N=64 bits in a javascript implementation relatively quickly, but for N=96, it really slows down as Pollard's Rho starts with large factors. This means my fast-and-loose assumption that "a constant number of iterations of Pollard Rho would work" isn't actually true!

Ok, chatting with Claude and I found a way to salvage the approach! (Also, it's apparently in the paper that's referenced in the post, kinda cool that my sketchy algorithm got halfway to the published result).

Basically, replace the Pollard Rho partial factorization with a method of Kalai [1] for generating random numbers _together with their prime factors_.

I'm able to run this at about 30 samples per second at 160bits, giving an estimate of ~14.1% of 160-bit numbers factoring into two 80-bit numbers.

[1]: https://link.springer.com/article/10.1007/s00145-003-0051-5

I think by "have 32 bit factors" the GP actually means "has a factorization as a product of two 32-bit numbers", rather than the natural meaning that doesn't work, as you pointed out.

There is a cute argument (I think it is due to Erdos) that, asymptotically, 0% of the integers in [0,n^2] appears in the "n by n multiplication table":

By Erdos-Kac, almost all integers of size about n^2 have about log(log(n^2)) ~ log(log(n)) prime factors. However, almost all integers in the multiplication table have about 2*log(log(n)) prime factors.

Kevin Ford gets much more precise asymptotic estimates.

They address this argument in the blog.

Where?

The mathematical term for this is the probability of a number being b-smooth. Here ‘b’ is 2^32

Related but not strong enough. 17 x 17 x 17 = 4,913 is 2^8-smooth - no prime factors larger than 2^8 - and it is less than 2^16, but 17 x 17 = 289 does not fit into a byte. Smoothness is required but not sufficient for a product representation to exist.

It's related, but not the same thing. For example, for b=10, the number 70=2x5x7 is b-smooth, but it cannot be written as the product of two numbers less than b. Here are the other b-smooth (counter)examples for b=10:

    | n   | factorization  | products of two numbers
    |-----|----------------|------------------------------------
    | 50  | 2 * 5^2        | 1x50, 2x25, 5x10
    | 60  | 2^2 * 3 * 5    | 1x60, 2x30, 3x20, 4x15, 5x12, 6x10
    | 70  | 2 * 5 * 7      | 1x70, 2x35, 5x14, 7x10
    | 75  | 3 * 5^2        | 1x75, 3x25, 5x15
    | 80  | 2^4 * 5        | 1x80, 2x40, 4x20, 5x16, 8x10
    | 84  | 2^2 * 3 * 7    | 1x84, 2x42, 3x28, 4x21, 6x14, 7x12
    | 90  | 2 * 3^2 * 5    | 1x90, 2x45, 3x30, 5x18, 6x15, 9x10
    | 96  | 2^5 * 3        | 1x96, 2x48, 3x32, 4x24, 6x16, 8x12
    | 98  | 2 * 7^2        | 1x98, 2x49, 7x14

This is something I had thought about some time back where I was thinking about the feasibility of somehow using the upper and lower registers inside a multiplier as general purpose storage for fun / seeing if you could make them more compact.

Anyway here is a fun pattern you get when you multiply 8 bit unsigned integers. Not all pairs of (upper bits, lower bits) are reachable, and it has a lot of distinct patterns.

https://i.imgur.com/Gb3HDR0.png

(Should I host the image on GitHub Gists so it doesn't vanish?)

This just seems like an expansion of prime numbers to includes factors in the 2^33+ range. Basically you're calculating if a number is prime but stopping the check when the factors go above 2^32.

Well, technically yes, but 'stopping the factors at 32 bits' is a plenty interesting constraint because it excludes all 64 bit composite numbers that have at least one factor above 2^32.

You have to redo the math to make the constraint work.

Having a prime factor greater than 2^32 accounts for about 80% of the 64-bit integers that can’t be expressed as a product of 32-bit integers. But it’s not the only way; you can also have three prime factors in the range (2^16, 2^32), for instance.

More precisely one in the range (a,2^32) and two in the range (2^32/a, 2^32). But if the latter have many duplicate prime factors it's worse.

Sure. You're just experiencing aliasing though. Are you not?

> There are 3,215,709,724,700,470,902 64-bit (unsigned) integers that can be written as a product of two 32-bit integers.

That can be written as a product of one or more pairs of 32 bit integers. So this is just not a bijective map.

I don't think I needed an AI-generated infographic of the headline. It looks like a product sales pitch.

Extremely strange way to deliver the headline (right before delivering the headline).

So you're better of using a 8x8->16 widening multiplication SIMD instruction or even just a multi register TBL/TBX instruction?

The way the headline is phrased doesn't really surprise me much. There aren't twice as many 64-bit integers than 32-bit ones; there are twice as many 33-bit integers as 32 bit ones, and there are 2^32 times as many 64-bit ones than 32-bit ones. It's like asking how many numbers between 1-64 you can get by multiplying numbers between 1-8; I think it's readily apparent that a pretty large portion are missed.

But you're using two 32-bit numbers, which have the same total bits as a 64-bit number. There are equally many 32-bit x 32-bit pairs as there are 64-bit numbers.

And there are as many pairs of numbers between 1-8 and numbers from 1-64, but it's still pretty apparent that most of them are not represented in the set of products.

Well, that is entirely not surprising. Pretty sure people writing not terrible hash functions figured it decades ago

I must be missing something. Aren’t ~50% of 64-bit integers the product of the number 2 and another 32-bit integer?

I don’t think so, because that only gets you up to 2x2^32, which is nowhere near halfway to 2^64

Going from 32 bits to 64 bits doesn't double the range (that would be adding 1 bit), it squares the range.

No. 50% of them are the product of 2 and a 63-bit integer.

Only 20% of 2 digit numbers are product of one digit numbers: 10 (1+9), 11, 12, 13 … 18 (9+9)

I wonder what kind of result you'd get with floats

Does it actually matter for hash uniformity, though?

Honestly, this is a larger portion than I would have expected.

If this seems counterintuitive, consider that only about a third of the two-digit numbers ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 63, 64, 72, 81}) can be written as the product of two one-digit numbers.

where is the graph and the theorem for integers of n bits, with n going to infinity?

The math was linked in the article

https://arxiv.org/pdf/1908.04251

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